What is the Poincare dual of a symplectic form? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:26:20Z http://mathoverflow.net/feeds/question/42331 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42331/what-is-the-poincare-dual-of-a-symplectic-form What is the Poincare dual of a symplectic form? Paul Siegel 2010-10-15T21:35:29Z 2010-10-23T14:32:03Z <p>Every symplectic form on a manifold $M^n$ determines a De Rham cohomology class in $H^2(M)$ (often a nontrivial class), and this in turn determines a class in $H_{n-2}(M)$. What in general can be said about this class? For example, over the rationals this class is represented by a submanifold of $M$; is it possible to explicitly describe such a submanifold in terms of the symplectic structure?</p> <p>If there is a nice answer to this question, does it also shed light on the Poincare duals of $\omega^2$, $\omega^3$, etc?</p> http://mathoverflow.net/questions/42331/what-is-the-poincare-dual-of-a-symplectic-form/42334#42334 Answer by Tim Perutz for What is the Poincare dual of a symplectic form? Tim Perutz 2010-10-15T22:30:34Z 2010-10-15T22:30:34Z <p>One of the big advances in symplectic topology in the 90s was Donaldson's theorem that when the symplectic class is integral, high multiples of its dual are represented by symplectic submanifolds. </p> <p>These submanifolds behave like hyperplane sections in algebraic geometry; for instance, they satisfy the Lefschetz hyperplane theorem. They form the fibres of "symplectic Lefschetz pencils". Their intersections can be made to give symplectic submanifolds dual to multiples of wedge-powers of $\omega$. </p> <p>Imagine first that $M$ is a compact complex manifold, $L\to M$ a hermitian, holomorphic line bundle, whose Chern connection has curvature $-2\pi i\omega$, a closed $(1,1)$-form. Then the zero-set of a $C^\infty$ section $s$ of $L^{\times k}$, if cut out transversely, is dual to $k[\omega]$. If $\omega$ is positive, the Kodaira embedding theorem then tells us that $L$ is ample: its high powers have enough holomorphic sections to embed $M$ into projective space. If $M$ is merely symplectic, with $-2\pi i\omega$ the curvature of some unitary connection in a hermitian line bundle, we can choose an almost complex structure $J$ on $M$ and consider transverse sections $s_k$ of $L^{\otimes k}$ for which, asymptotically, the $(0,1)$-part of $\nabla s_k$ along $s_k^{-1}(0)$ is much smaller than the $(1,0)$-part. Then $s_k^{-1}(0)$ will not quite be a $J$-holomorphic submanifold, but for $k \gg 0$ its tangent spaces will be close enough to being $J$-linear that it will still be a symplectic submanifold. </p> <p>References: S. K. Donaldson, "Symplectic submanifolds and almost-complex geometry", J. Differential Geom. Volume 44, Number 4 (1996), 666-705; "Lefschetz pencils on symplectic manifolds", J. Differential Geom. Volume 53, Number 2 (1999), 205-236. </p> <p>These papers are brilliant both geometrically and analytically: the analysis is mostly low-tech but extremely subtle.</p> http://mathoverflow.net/questions/42331/what-is-the-poincare-dual-of-a-symplectic-form/43303#43303 Answer by SB for What is the Poincare dual of a symplectic form? SB 2010-10-23T14:32:03Z 2010-10-23T14:32:03Z <p>Paul Biran has also studied this situation where $(M, \omega, J)$ is Kahler, $[\omega] \in H^2(M, \mathbb{Z})$ and $\Sigma$ is a complex hypersurface Poincare dual to $k[\omega]$. He has proved that in this setting, $M$ can be decomposed symplectically as the unit normal disk bundle of $\Sigma$ and an isotropic CW-complex.</p> <p>This decomposition result is then used to investigate various embedding questions into $M$.</p> <p>Biran: "Lagrangian barriers and symplectic embeddings. Geom. Funct. Anal. 11 (2001), no. 3, 407–464"</p> <p>Biran and Cieliebak: "Symplectic topology on subcritical manifolds. Comment. Math. Helv. 76 (2001), no. 4, 712–753"</p> <p>Biran: "Lagrangian non-intersections. Geom. Funct. Anal. 16 (2006), no. 2, 279–326"</p>